71Ans (a)I did it practically(though instead of a bird I used a stone), & I experienced that when I was holding it with the stone kept the at the base of cage it weighs more than when I hold it with the stone kept hung by a thread inside the cage without touching the walls of the cage. So, my conclusion is that I had to use less energy,i.e, less effort to hold it.
I don't think you have done the experiment correctly. Cause as it mentions in the question No. 8 (a), the bird flies and remains in air so it no longer a part of the The Person and Cage System. But you have hung the stone with a thread which in turn creates a tension and certainly acts same as if the stone was kept within it.
Please let me know If I'm wrong Cause I'm not an expert at all!
11My final verdict is that when the bird will be flying in the cage without touching any of the
walls of the cage, let the cage have solid or grilled base, we will have to apply less amount of
force to lift it
and for part (b) you know the answer very well.
:)
11
Khyati
·2010-09-05 11:14:15
the base is only solid one but the rest cage is grilled so wont the air will go out from there??
Air requires the space to get out and its getting from other part of the cage, its not necessary that it will swooshed only through the base, it can go out from any where so the force that is acting downward is only Mg, so we have to less force to lift the cage.
I have mention the completely close case only in part (b) so in the part (a) the cage is not completely closed and hence air is getting the space to get out.
71
Vivek @ Born this Way
·2010-09-05 08:13:18
in case when the bird is sitting quietly, the net force exerted downward is mg + Mg in both cases
[m=mass of bird and M=mass of cage]
now suppose the bird is flying,
the bird feels weightless...since its weight mg is balanced by an upthrust provided by moving it's wings
now the bird is thrusting the air downwards and in (I) the air strikes the solid base of the cage with a force mg so net downward force is mg(downthrust)+Mg(weight of cage)+0(weight of bird)
so to lift the cage same amt of force is reqd.
now had the base been grilled one,
the thrusted air would have swooshed through the cage and hit the ground....
thus net downward force is Mg only.....
thus less force is reqd. to lift the cage...
11
Khyati
·2010-09-05 07:48:07
what answer will be different? Please be clear and explain in detail.
71
Vivek @ Born this Way
·2010-09-05 06:48:54
He has said that In both cases answers aren't same. Grilled one will have less weight than that one which is closed!
11
Khyati
·2010-09-05 04:54:28
you say what's the difference in using the solid base and a grilled base, in both the cases the answer gonna be same
11
Khyati
·2010-09-05 04:51:06
What you didn't get, its the same as I was saying that when the bird is flying we have to apply lesser amount of force to lift the cage.
This time jara deatil mein explain kiya hai ki aur kya kya hota hai :P
49so??
couldnt understand what u trying to say??
clarify!!
11
Khyati
·2010-09-05 00:20:09
8 a) When the bird flies inside the cage without touching the walls, then you have only the
weight of mg to bear .. This is with the assumption that the cage allows the air to pass out
of it. The explanation is simple.. The bird in order to fly flaps its wings and pushes the air
down and as a result lifts up. The air pushed down by the flapping increases the force on the
hand carrying it from mg to a little higher.. But, this air escapes through the bars and doesn't
create the maximum amount of force on the hand. (It's true that as this air goes out, new air
comes in with almost the same velocity but the resistance between air particles reduces the
amount of force it can generate with the distance it has to move)
11
Khyati
·2010-09-05 00:10:23
@ Vivek I didn't tied the string to the cage, I was holding it with my hands so it was as if the bird was flying in the cage without touching any wall of the cage.
49
Subhomoy Bakshi
·2010-09-04 10:24:52
8)
well lets first of all divide the problem into 2 parts....
(I) when the cage has a solid base
and
(II)When the base is a grill
in case when the bird is sitting quietly, the net force exerted downward is mg + Mg in both cases
[m=mass of bird and M=mass of cage]
now suppose the bird is flying,
the bird feels weightless...since its weight mg is balanced by an upthrust provided by moving it's wings
now the bird is thrusting the air downwards and in (I) the air strikes the solid base of the cage with a force mg so net downward force is mg(downthrust)+Mg(weight of cage)+0(weight of bird)
so to lift the cage same amt of force is reqd.
now had the base been grilled one,
the thrusted air would have swooshed through the cage and hit the ground....
thus net downward force is Mg only.....
thus less force is reqd. to lift the cage...
1For Q-3, in case of perfectly elastic collision, the velocities of two bodies, (I think), should be same at the instant of maximum deformation (or minimum Kinetic Energy of the system) of the two bodies..... where am i wrong???
71
Vivek @ Born this Way
·2010-09-03 23:18:21
For the 8(b); Unless we are changing the mass to the body, I don't think there should be any change!
621) Internal forces can change the linear momentum of portion of the system, but they can't
change the the total linear momentum of the entire system. WHY?
The reason is that the force is rate of change of momentum...
so we have F1 acting on one part and -F1 on another.
So we have the change of momentum on one part being counter balanced by the change on another...
Hence the net change in momentum of the system as a whole is zero...
62
Lokesh Verma
·2010-08-07 12:34:41
6)To accelerate the car we ignite petrol in the engine of the car. Since only an external force can accelerate the car is it proper to say that "the force generated by the engine accelerates the car?
no not at all... the engine is a part of the car...
if you observe carefully the engine does not move the car... it creates a torque.. which causes the rotation of the wheels of the car.. this leads to friction between the tyre and ground.. which is a force external to the car which leads to the motion of the car.
think of the situation when there is not friction at all.. will the car move?
&*oops din see subhomoy's reply
11
Khyati
·2010-08-07 12:33:45
@ subhomoy
But in 7th one it is clearly mentioned that the collision is ELASTIC for which e = 1
& e≠1 for inelastic collision
62
Lokesh Verma
·2010-08-07 12:32:14
for 7, your logic is correct...
m1v1+m2v2=m1v2+m2v1
now we have
(m1-m2)(v1-v2)=0
where v1 and v2 are vectors..
hence either m1=m2 or v1=v2
but if v1=v2 collision cant take place :P
hence your logic is correct.
49
Subhomoy Bakshi
·2010-08-07 12:32:13
6) actually it is the friction that drives the car!!
the internal forces just rotates the wheels!!
and torque can be provided by internal forces! :)
49
Subhomoy Bakshi
·2010-08-07 12:30:09
6) well we can always say "the power generated in the engine ensures motion"!!
49
Subhomoy Bakshi
·2010-08-07 12:26:47
4)Law of momentum conservation will be valid and that is
∫F.dt = change in momentum (Newton's 2nd law!!!)
11
Khyati
·2010-08-07 12:15:00
One more question
8) You are holding a cage containing a bird.
(a) Do you have to make less effort if the bird flies from its position and manages to stay in the middle without touching the walls of cage ?
(b) Does it makes difference whether the cage is completely closed or it has rods to let air pass?
Ans (a)I did it practically(though instead of a bird I used a stone), & I experienced that when I was holding it with the stone kept the at the base of cage it weighs more than when I hold it with the stone kept hung by a thread inside the cage without touching the walls of the cage. So, my conclusion is that I had to use less energy,i.e, less effort to hold it.
(b)I think if the cage is closed then, I'll have to apply the same force to hold the cage whether the bird is in air or in its position because it would be a part of the closed system where no mass is allowed to leave or enter the system (Hope I am correct in assuming the completely closed cage as a Closed system :))
Please correct me if I am wrong
11
Khyati
·2010-08-07 10:37:44
Thanks a lot Nishant Sir :) :)
11
Khyati
·2010-08-07 10:37:02
5)Two bodies make an elastic head on collision on a smooth horizontal table kept in a car. Will there be a change i the results if the car is accelerating on a horizontal road because of the non inertial character of the frame? Will the conservation of linear momentum will be valid here?
Same thoughts as I have for question 4
6)To accelerate the car we ignite petrol in the engine of the car. Since only an external force can accelerate the car is it proper to say that "the force generated by the engine accelerates the car?
Completely confused. No thoughts at all :P :P
11
Khyati
·2010-08-07 10:33:35
7)In one dimensional elastic collision of equal masses the velocities are interchanged. Can velocities in a one dimensional collisions be interchanged if masses are not equal?
Mine answer in NO for this, because it will disobey the law of conservation of linear momentum
4)A collision experiment is done on a horizontal table kept in an elevator. will there be any change in the results if the elevator is accelerating up or down because of non inertial frame of reference??
I think it will surely affect the collision experiment because since the elevator in accelerating that will surely affect the motion of the particles taken for the experiment. They will fall down from the table due the acceleration of the elevator. Since it is acceleration it implies some external force is action on it, so here the law of conservation of linear momentum won't be valid
62
Lokesh Verma
·2010-08-06 21:46:12
give your justification to a few remaining questions (or any thing that you think correct and we will try to comment or improve on your thinking)
I feel otherwise it will not help you understand things as well as you should [1]
62
Lokesh Verma
·2010-08-06 21:44:18
3)In a head on collision between two particles is it necessary that the particles will acquire a common velocity at least for one instant? Is it the instant just after the collision has taken place........
Yes and no..
See in case of a perfectly elastic collision.. there will be no such instant.. the change takes immediately..
but in case of a "real world" collision what you have said is correct... because there will be a retardation ni the speed of one body and accelerataion in the other...
62
Lokesh Verma
·2010-08-06 21:41:48
2)How can the centre of mass of a body be at a point outside the body, though we call it a point at which the whole mass of the body is assumed to be concentrated? Is it mere a assumption or does it has a real existence too??
Think of a hollow sphere... the center of mass is outside the part containing any mass.... (ie the interior)
or take a ring..
It is an assumption that we can use a point mass instead of a large body for one dimensional motion. The CM of a body is real! in the sense that it exists and we have the coordinates which give that point.....